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Miscellaneous Exercise · Q13

Q.Integrate the function ex(1+ex)(2+ex)\frac{e^x}{(1+e^x)(2+e^x)}

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The integral ∫ex(1+ex)(2+ex) dx\int \frac{e^x}{(1+e^x)(2+e^x)} \, dx is solved by substituting t=ext = e^x, then applying partial fraction decomposition to the resulting rational function. The final answer is log⁡∣1+ex2+ex∣+C\boxed{\log\left|\frac{1+e^x}{2+e^x}\right| + C}.

Why Partial Fractions Work Here

When you see a product of linear factors in the denominator — like (1+ex)(2+ex)(1+e^x)(2+e^x) — and a numerator that is essentially the derivative of one of those factors, your first instinct should be substitution. Here, exe^x is both the numerator and the derivative of exe^x itself. That’s a strong hint: let t=ext = e^x, so dt=exdxdt = e^x dx, and the integral becomes a clean rational function in tt.

The denominator becomes (1+t)(2+t)(1+t)(2+t), and the numerator is just dtdt. So we’re integrating 1(1+t)(2+t) dt\frac{1}{(1+t)(2+t)}\, dt. This is a textbook partial fractions problem: split the fraction into two simpler pieces, each of which integrates to a logarithm.

Step-by-Step Solution

1. Substitute t=ext = e^x

Let t=ext = e^x. Then dt=exdxdt = e^x dx, which is exactly the numerator of our integrand. So:

∫ex(1+ex)(2+ex) dx=∫1(1+t)(2+t) dt\int \frac{e^x}{(1+e^x)(2+e^x)} \, dx = \int \frac{1}{(1+t)(2+t)} \, dt

Tip

The substitution t=ext = e^x is natural here because exe^x appears both in the numerator and inside the denominator factors. Always look for a function and its derivative when choosing a substitution.

2. Set up partial fractions

We want to write:

1(1+t)(2+t)=A1+t+B2+t\frac{1}{(1+t)(2+t)} = \frac{A}{1+t} + \frac{B}{2+t}

Multiply both sides by (1+t)(2+t)(1+t)(2+t):

1=A(2+t)+B(1+t)1 = A(2+t) + B(1+t)

3. Solve for AA and BB

We can solve by choosing convenient values of tt:

  • Let t=−1t = -1: then 1=A(2−1)+B(0)  ⟹  1=A⋅1  ⟹  A=11 = A(2-1) + B(0) \implies 1 = A \cdot 1 \implies A = 1
  • Let t=−2t = -2: then 1=A(0)+B(1−2)  ⟹  1=B⋅(−1)  ⟹  B=−11 = A(0) + B(1-2) \implies 1 = B \cdot (-1) \implies B = -1
Watch out

A common mistake is to forget the sign when solving for BB. Double-check: plugging t=−2t = -2 gives 1=B(−1)1 = B(-1), so B=−1B = -1, not +1+1.

4. Rewrite the integral

Now we have:

∫1(1+t)(2+t) dt=∫(11+t−12+t)dt\int \frac{1}{(1+t)(2+t)} \, dt = \int \left( \frac{1}{1+t} - \frac{1}{2+t} \right) dt

5. Integrate term by term

Each term integrates to a natural logarithm: …

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